if 2 cos (A+B) = 1 and 2 sin (A-B) = 1 then the value of A and B are. (answer is 45⁰,15⁰) how❓
Answers
Step-by-step explanation:
Given :-
2 cos (A+B) = 1
2 sin (A-B) = 1
To find :-
Find the values of A and B ?
Solution :-
Given that
2 cos (A+B) = 1
=> Cos (A+B) = 1/2
=> Cos (A+B) = Cos 60°
=> A+B = 60° -----------------(1)
and
2 sin (A-B) = 1
=> Sin (A-B) = 1/2
=> Sin (A-B) = Sin 30°
=>A-B = 30° ----------------(2)
On adding (1)&(2) then
A + B = 60°
A - B = 30°
(-) (+) (-)
____________
0 + 2B = 30°
____________
=> 2B = 30°
=> B = 30°/2
=> B = 15°
On substituting the value of B in (1) then
=> A + 15° = 60°
=> A = 60° - 15°
=> A = 45°
Therefore, A = 45° and B = 15°
Answer:-
The value of A = 45°
The value of B = 15°
Check :-
If A = 45° and B = 15° then
Equation (1):-
LHS = 2 cos (45°+15°)
=> 2 Cos 60°
=> 2×1/2
=> 2/2
=> 1
LHS = RHS is true for A = 45° and B = 15°
Equation (2):-
LHS = 2 sin (A-B)
=> 2 Sin (45°-15°)
=> 2 Sin 30°
=> 2(1/2)
=> 2/2
=> 1
=> RHS
LHS = RHS is true for A = 45° and B = 15°
Verified the given relations in the given problem.
Used formulae:-
- Cos 60° = 1/2
- Sin 30° = 1/2